Growth conditions are imposed on f such that the following boundary value problem: (−1)my(2m) = f(t, y), αi+1y(2i)(0) − βi+1y(2i+1)(0) = γi+1y(2i)(1) + δi+1y(2i+1)(1) = 0, 0 ≤ i ≤ m−1, has an arbitrary number of positive solutions.
關聯:
Computers and Mathematics With Applications 40(2-3), pp.231-237
